JEE MAIN Definite Integration Previous Year Questions (PYQs) – Page 12 of 17
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The area bounded by the curve 4y2 = x2(4 $-$ x)(x $-$ 2) is equal to :
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Solution
Let $[x]$ denote the greatest integer $\le x$. Consider the function
$$f(x)=\max\{x^{2},\,1+[x]\}.$$
Then the value of the integral $\displaystyle \int_{0}^{2} f(x)\,dx$ i
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Solution
If $\displaystyle \int_{0}^{1} \frac{1}{(5+2x-2x^2)\,(1+e^{\,2-4x})}\,dx=\frac{1}{\alpha}\log_e\!\left(\frac{\alpha+1}{\beta}\right),\ \alpha,\beta>0,$ then $\alpha^4-\beta^4$ is equal to:
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Solution
The integral $\displaystyle \int_{1/4}^{3/4} \cos\left( 2\cot^{-1}\sqrt{\frac{1-x}{1+x}} \right),dx$ is equal to:
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Solution
If the value of the integral $\int\limits_0^{{1 \over 2}} {{{{x^2}} \over {{{\left( {1 - {x^2}} \right)}^{{3 \over 2}}}}}} dx$ is ${k \over 6}$, then k is equal to :
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Solution
Let g(x) = $\int_0^x {f(t)dt} $, where f is continuous function in [ 0, 3 ] such that ${1 \over 3}$ $ \le $ f(t) $ \le $ 1 for all t$\in$ [0, 1] and 0 $ \le $ f(t) $ \le $ ${1 \over 2}$ for all t$\in$ (1, 3]. The largest possible interval in which g(3) lies is :
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Solution
The value of the integral
$\displaystyle \int_{-1}^{2} \log_e \big(x + \sqrt{x^2 + 1}\big),dx$ is
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Solution
Let a be a positive real number such that $\int_0^a {{e^{x - [x]}}} dx = 10e - 9$ where [ x ] is the greatest integer less than or equal to x. Then a is equal to:
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Solution
$\displaystyle \int e^{\sec x},\big(\sec x\tan x,f(x)+\sec x\tan x+\sec^{2}x\big),dx ;=; e^{\sec x}f(x)+C$
Then a possible choice of $f(x)$ is:
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Solution
The value of the integral $\int\limits_{ - 1}^1 {{{\log }_e}(\sqrt {1 - x} + \sqrt {1 + x} )dx} $ is equal to:
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Solution
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